Suppose $Q(x):[0,1]\to[0,1]$ is a segment of a convex set which is concave downwards and locally Lipchitz and differentiable a.e. such that $Q(0)=Q(1)=0$
Is $Q''(x)$ going to be tending to $-\infty$ at only finite points?
My first thought is yes as if the second derivative is too negative too often the constraint that $Q(1)=0$ will be violated. However, is it possible to make them summable?
The answer is no. Consider the function $$ u(x) = \sum_{j=1}^\infty \frac{-2^{-j}}{\sqrt{|x - 2^{-j}|}} $$ which clearly is integrable, negative, and has poles at $2^{-j}$, i.e. at countably many points in $[0,1]$. Then set $$ Q_0(x) = \int_0^x \int_0^t u(s) \, ds \, dt, \quad Q(x) = Q_0(x) - Q_0(1)x \, . $$
Then $Q'' = u \le 0$ and $Q'$ is continuous, hence $Q$ is even differentiable everywhere.